Page 390 - Engineering Electromagnetics, 8th Edition

372 ENGINEERING ELECTROMAGNETICS

It is evident that taking the partial derivative of any field quantity with respect
to time is equivalent to multiplying the corresponding phasor by jω.Asanexample,
we can express Eq. (8) (using sinusoidal fields) as

∂H y ∂E x
=− 0 (20)
∂z ∂t
where, in a manner consistent with (19):
1 1
E x (z, t) = E xs (z) e jωt + c.c. and H y (z, t) = H ys (z) e jωt + c.c. (21)
2 2
On substituting the fields in (21) into (20), the latter equation simplifies to

dH ys (z) =− jω 0 E xs (z) (22)
dz
In obtaining this equation, we note first that the complex conjugate terms in (21)
produce their own separate equation, redundant with (22); second, the e jωt factors,
common to both sides, have divided out; third, the partial derivative with z becomes
the total derivative, since the phasor, H ys , depends only on z.
We next apply this result to Maxwell’s equations, to obtain them in phasor form.
Substituting the field as expressed in (21) into Eqs. (1) through (4) results in

∇× H s = jω 0 E s (23)
∇× E s =− jωµ 0 H s (24)
∇ · E s = 0 (25)
∇ · H s = 0 (26)

It should be noted that (25) and (26) are no longer independent relationships, for they
can be obtained by taking the divergence of (23) and (24), respectively.
Eqs. (23) through (26) may be used to obtain the sinusoidal steady-state vector
form of the wave equation in free space. We begin by taking the curl of both sides
of (24):
2
∇× ∇× E s =− jωµ 0 ∇× H s =∇(∇ · E s ) −∇ E s (27)

where the last equality is an identity, which defines the vector Laplacian of E s :
2
∇ E s =∇(∇ · E s ) −∇ ×∇ × E s

From (25), we note that ∇ · E s = 0. Using this, and substituting (23) in (27), we
obtain

2 2 (28)
∇ E s =−k E s
0